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INPhO 2026, Теория — Задача 5. From Kepler’s archive

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Всички материали от INPhO 2026

Indian National Physics Olympiad (INPhO)-2026 · Question Paper · 1 февруари 2026 г. · 18 т.

Внимание: Бележка към темата

Matching original question and official key booklets are retained in full. The separate blank Answer Booklet, Summary Answersheet copies and blank graph paper referred to in the questions are absent from the archive; the supplied completed Kepler/glass answer-sheet illustrations remain in the key. Native question pages7 and9 are blank. Question6 says photographs on the next page although its photos occur on native p10 after blank p9. Printed source errors/conventions are preserved and attributed locally: P2 negative dM followed by positive boxed dUm and changing absolute-work notation; P3 Qa=-W followed by positive released heat, dropped-mass wording and P0/V0 notation; P4 derivative/height signs, c/cs and time-delay description; P5 intermediate phiP versus final phiC, log log and r/logarithm numerical scales. The graph/grid pixels and printed fitted results are source data, not newly calculated measurements.

Question booklet · Instructions and Table of Constants

Indian National Physics Olympiad (INPhO)-2026

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION · Tata Institute of Fundamental Research · V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Question Paper

Date: 01 February 2026 · Time: 09:00–12:00 (3 hours) · Maximum Marks: 75

Roll Number: □□□□ – □□□□ – □□□□

Instructions

This booklet consists of 10 pages and total of 6 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Take as many data points as possible for the analysis.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.
Except for writing your roll number, no rough work or scribbling is allowed on the question paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$
One Atmospheric pressure	atm	$1.013\times10^5\ \mathrm{Pa}$

Question Number	1	2	3	4	5	6	Total
Maximum Marks	12	12	11	10	18	12	75

Solution booklet · Instructions, constants and alternate-method note

Indian National Physics Olympiad (INPhO)-2026

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION · Tata Institute of Fundamental Research · V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Question Paper

Date: 01 February 2026 · Time: 09:00–12:00 (3 hours) · Maximum Marks: 75

Roll Number: □□□□ – □□□□ – □□□□

Instructions

This booklet consists of 21 pages and total of 6 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Take as many data points as possible for the analysis.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.
Except for writing your roll number, no rough work or scribbling is allowed on the question paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$
One Atmospheric pressure	atm	$1.013\times10^5\ \mathrm{Pa}$

Question Number	1	2	3	4	5	6	Total
Maximum Marks	12	12	11	10	18	12	75

Please note that alternate/equivalent methods and different ways of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

A short note found in Kepler’s archive describes a curious central-force problem. The note states that the radial potential has the form $U(r)=kr^n$ , where $k>0$ is a dimensional constant, $n$ is a positive integer, and $r$ is the distance from a fixed origin. Kepler also recorded the particle’s precise trajectory by listing its $x$ - $y$ coordinates in its plane of motion (see the figure below). His sketch includes several short tangent segments drawn at selected points, marked by $\times$ .

Whole native Kepler trajectory and tangent segments, every marked red point including P, dashed origin axes, complete original numerical scales and x/y labels. Lossless clockwise90° pixel rotation for reading.

The coordinates $x$ and $y$ are given in arbitrary units. Kepler’s notes indicate that at point P, the kinetic energy is exactly one quarter of the total mechanical energy. He further noted that the exponent $n$ could be determined by performing calculations based on the graph and by constructing a linear plot. Unfortunately, the remainder of the manuscript explaining this method has been lost.

To understand what Kepler did, we define a few variables below. The figure below shows the trajectory of a particle moving under a central force.

Native central-force geometry showing C,O,velocity and radial/tangential components and angle phiC; native x/y axis labels retained without redraw.

Consider a fixed point C on the trajectory (shown by $\times$ ), located at a distance $r_C$ from the origin O. At point C, let the speed of the particle be $v_C$ , and let the radial and tangential components of its velocity be $v_{rC}$ and $v_{\phi C}$ , respectively ( $\vec v_{\phi C}$ is $\perp$ to $\vec v_{rC}$ ).

The angle between $\vec v_{\phi C}$ and the velocity vector $\vec v_C$ is denoted by $\phi_C$ .

At an arbitrary point on the trajectory, the particle is at a distance $r$ from the origin, its speed is $v$ , and the corresponding angle between $\vec v$ and its tangential component $\vec v_\phi$ is $\phi$ .

(a) The speed $v$ at an arbitrary point can be written in terms of the speed at point C as $v=\alpha v_C$ . Express $\alpha$ in terms of $r_C$ , $r$ , $\phi_C$ , and $\phi$ . [5 т.]

(b) Two versions of Kepler’s diagram are given in the answer sheet: one with tangents drawn and one without. You may use either or both figures as needed.

Use these to devise a method to find the value of $n$ . Perform all relevant analyses using the figures provided on the answer sheet. Finally, use the graph paper at the end of the answer sheet for plotting a linear graph to determine $n$ , and report any necessary data tables in the detailed answersheet. [13 т.]

Отговори

Решение

(a)

From the given trajectory,

v_r=v_\phi\tan\phi,\qquad v_\phi=v\cos\phi.

In central forces, the angular momentum $l$ is conserved.

l=rmv_\phi=\text{constant}.

Using conservation of momentum at point C and any arbitrary point, we get

rv_\phi=r_Cv_{\phi C},

rv\cos\phi=r_Cv_C\cos\phi_P,

v=v_C\frac{r_C\cos\phi_P}{r\cos\phi},\qquad v=\alpha v_C,

where

\boxed{\alpha=\frac{r_C\cos\phi_C}{r\cos\phi}.}

(b)

From the previous part $v=\alpha v_P$ .

It is given that kinetic energy at point P is 1/4th of the total energy $E$ .

\frac12 mv_P^2=\frac E4,

2mv_P^2=E=kr^n+\frac12 mv^2,

4mv_P^2=2kr^n+m\alpha^2v_P^2,

mv_P^2(4-\alpha^2)=2kr^n.

Taking log log of both the sides and rearranging

\boxed{\ln(4-\alpha^2)=n\ln r+\ln\frac{2k}{mv_P^2}.}

The slope of graph plotted between $\ln(4-\alpha^2)$ vs $\ln r$ can be used to calculate the value of $n$ . The trajectory of the particle is given in the figure, the tangents are drawn at some points. We measure $r$ and $\phi$ at those points using a ruler and protractor. Also measure $r_C$ , $\phi_C$ . We use these values to calculate $\alpha$ for each $\times$ on the trajectory. In the figure on the next page, the angles measured are $90^{\circ}-\phi$ . The final data is tabulated on the next page.

Whole native completed Summary Answer sheet trajectory with measured angles, numbered points1–8 and P, full axes and all handwritten construction lines. Lossless clockwise90° rotation.

Sr.No	$r$ (in arbitrary units)	$\phi$ (in degrees)	$\ln(r)$	$\ln(4-\alpha^2)$
1	5.2	41	-2.957	0.107
2	4.7	36	-3.058	-0.079
3	6.7	40	-2.703	0.838
4	5.8	43	-2.847	0.424
5	3.5	2	-3.352	-1.008
6	6.2	43	-2.781	0.608
7	4.1	28	-3.194	-0.5
8	5.7	43	-2.865	0.365
P	7.3	24

Complete native handwritten Kepler log plot, green grid, all points, full axes/scales, slope construction and 2.764 annotation; original Detailed Answer sheet header retained. Lossless clockwise90° rotation.

The slope of the graph is 2.76. so the value of $\boxed{n\text{ is }3}$ . We rule out $n=2$ case, since for that, the trajectory will be an ellipse.

Source note: the two intermediate (a) expressions print $\phi_P$ while the final boxed expression prints $\phi_C$ . The table’s printed $r$ values and negative $\ln(r)$ values use different numerical scales; both columns are retained verbatim. “log log” and the reference to $r_C,\phi_C$ in (b) are printed in the key.

Теми и трудност

Оценена трудност: 4/5.

Механика · Момент на импулса · Работа и енергия · Орбитално движение

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: INPhO2026-Question.pdf · официални решения: INPhO2026-Solution.pdf